$f(x) = x^{2}-2x+4(h(x))$ $h(x) = -2x$ $ h(f(2)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(2)$ . Then we'll know what to plug into the outer function. $f(2) = 2^{2}+(-2)(2)+4(h(2))$ To solve for the value of $f$ , we need to solve for the value of $h(2)$ $h(2) = (-2)(2)$ $h(2) = -4$ That means $f(2) = 2^{2}+(-2)(2)+(4)(-4)$ $f(2) = -16$ Now we know that $f(2) = -16$ . Let's solve for $h(f(2))$ , which is $h(-16)$ $h(-16) = (-2)(-16)$ $h(-16) = 32$